Question: The polynomial $P$ is graphed. ${1}$ ${2}$ ${3}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${\llap{-}2}$ ${\llap{-}3}$ $y$ $x$ $ P$ Graph of cubic polynomial P that passes through the points: negative 2, negative 3; 0, 3; and 3, 1. What is the remainder when $P(x)$ is divided by $(x-3)$ ?
We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $P(x)$ is divided by $(x-3)$, is $P({3})$. According to the graph, $P({3})=1$. ${1}$ ${2}$ ${3}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${\llap{-}2}$ ${\llap{-}3}$ $y$ $x$ $ P$ Same graph as above with point marked at 3, 1 In conclusion, the remainder when $P(x)$ is divided by $(x-3)$ is $1$.